Pulling a tablecloth from under a glass

[Leo Liu]

Homework Statement

This is a statement.

Relevant Equations

$f=\mu_{k} N$ $F=ma$

The question is stated above.

I tried to solve it on my own and I obtained the following three questions:
$$\begin{cases}
a_{cloth}=\frac{F-\mu mg} M\\
\\
a_{glass}=\frac{\mu mg}m=\mu g
\end{cases}$$
Where small m is the mass of the glass and capital M is the mass of the cloth. However, the first equation does not appear in the solution given by my answer book. So I think I must have wrongly interpreted the question. I would like someone to explain the question to me. Many thanks.
P.S. The official answer is shown below:

 

[hutchphd]

Your second equation should be $=\mu g$
The detailed acceleration of the cloth doesn't matter...only the time it takes. Presumably the "puller" can supply whatever force is required.

 

[Leo Liu]

Your second equation should be

It was a typo. Sorry.

The detailed acceleration of the cloth doesn't matter...only the time it takes. Presumably the "puller" can supply whatever force is required.

You have got the point. Thanks.

Update: Yet if I cross out the second equation, the time it takes to yank the cloth out is independent of the force. Since there is nothing to be maximized, what's the point of asking for the "longest time"?

 

[hutchphd]

What "force" are you solving for?

 

[etotheipi]

Where small m is the mass of the glass and capital M is the mass of the cloth. However, the second equation does not appear in the solution given by my answer book.

Surely it is the second equation that does appear in the given solution?

As @hutchphd alluded to, the dynamics of the cloth are irrelevant. We essentially have a glass that accelerates uniformly at $\mu g$ and then decelerates, in a symmetrical manner, with a magnitude of $\mu g$. The symmetry allows you to deduce the times for each stage are equal. The speed time graph is an isosceles triangle!

It is the longest time because the cloth can be pulled out arbitrarily quickly so that negligible impulse is initially imparted to the glass, causing it to come to rest well before the end of the table.

 

[TSny]

Can the answer T = 1/4 s be correct? If the cloth accelerates the glass for 1/4 s, where will the glass be located on the table at the instant the cloth is completely out from under the glass?

 

[hutchphd]

We essentially have a glass that accelerates uniformly at μg and then decelerates, in a symmetrical manner, with a magnitude of μg. The symmetry allows you to deduce the times for each stage are equal.

Not quite. I think the assumption is that without the cloth the friction is big so it decelerates "instantaneously"...makes the problem easier.too

 

[etotheipi]

Can the answer T = 1/4 s be correct? If the cloth accelerates the glass for 1/4 s, where will the glass be located on the table at the instant the cloth is completely out from under the glass?

Did they forget to add the distance traveled during the acceleration phase?

Not quite. I think the assumption is that without the cloth the friction is big so it decelerates "instantaneously"...makes the problem easier.too

I will bite the bullet... doesn't the question say the coefficient of friction without the cloth (i.e. glass on table) is 0.5?

 

[Leo Liu]

It is the longest time because the cloth can be pulled out arbitrarily quickly so that negligible impulse is imparted to the glass, causing it to come to rest well before the end of the table.

Oh this makes sense. So we only pull the tablecloth during the first half of the journey. After the external force is removed, the friction changes its direction. Thanks James!

Surely it is the second equation that does appear in the given solution?

Meh. It's another typo .

 

[TSny]

Did they forget to add the distance traveled during the acceleration phase?

They let $s_0$ represent the distance traveled during just the deceleration phase. But they took this distance to be 6".

 

[hutchphd]

I will bite the bullet... doesn't the question say the coefficient of friction without the cloth (i.e. glass on table) is 0.5?

Oops. You are indeed correct. Doesn't that make the published answer incorrect by 2?)

 

[etotheipi]

They let $s_0$ represent the distance traveled during just the deceleration phase. But they took this distance to be 6".

Oops. You are indeed correct. Doesn't that make the published answer incorrect by 2?)

Yeah, they should have $s = \mu g T^2 = 6\text{in}$

 

[haruspex]

Oops. You are indeed correct. Doesn't that make the published answer incorrect by 2?)

Reading the official solution, it looks like they left out the distance covered while being accelerated by the cloth.
It makes the the answer too much by a factor √2.